Solve the following Problem from Middle School Math Word Problems.

5 years ago, A was thrice as old as B and 10 years later, A shall be twice as old as B. What are the present ages of A and B ?

Solution to Example 1 of Middle School Math Word Problems :

Let x and y be the present ages of A and B respectively. 5 years ago their ages will be x - 5 for A and y - 5 for B. 5 years ago, A was thrice as old as B ⇒ x - 5 = 3(y - 5) = 3y - 15 ⇒ x - 3y = -15 + 5 = -10 ⇒ -x + 3y = 10....................................(i) 10 years later their ages will be x + 10 for A and y + 10 for B. 10 years later, A shall be twice as old as B ⇒ x + 10 = 2(y + 10) = 2y + 20 ⇒ x - 2y = 20 - 10 ⇒ x - 2y = 10....................................(ii)

Equations (i) and (ii) are the Linear Equations in two variables formed by converting the given word statements to the symbolic language.

Now we have to solve these simultaneous equations. (i) + (ii) gives 3y - 2y = 10 + 10 ⇒ y = 20. Using this in (ii), we get x - 2(20) = 10 ⇒ x -40 = 10 ⇒ x = 10 + 40 = 50. Thus A is 50 years old and B is 20 years old. Ans.

Check: 5 years ago, A's age = 50 - 5 = 45; B's age = 20 - 5 = 15 5 years ago, A's age = 45 = 3 x 15 = thrice B's age. (verified.) 10 years later, A's age = 50 + 10 = 60; B's age = 20 + 10 = 30 10 years later, A's age = 60 = 2 x 30 = twice B's age. (verified.)

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Solve the following problem from Middle School Math Word Problems.

The difference of the digits of a two-digit number is 3. If 4 times the number is equal to 7 times the number obtained by reversing the digits, find the number.

Solution to Example 2 of Middle School Math Word Problems

By data, 4 times the number = 7 times the number with the digits reversed. ⇒ the number with the digits reversed < the original number. ⇒ In the original number, ten's digit > unit's digit. Let x be the ten's digit and y be the unit's digit in the original number. Then x > y. Their difference is 3 means x - y = 3........................(i) The original number = y + 10x The number with the digits reversed = x + 10y 4 times the number = 7 times the number with the digits reversed. ⇒ 4(y + 10x) = 7(x + 10y) ⇒ 4y + 40x = 7x + 70y ⇒ 40x - 7x = 70y - 4y ⇒ 33x = 66y ⇒ x = 2y.......................(ii)

Equations (i) and (ii) are the Linear Equations in two variables formed by converting the given word statements to the symbolic language.

Now we have to solve these simultaneous equations. Using (ii) in (i), we get 2y - y = 3 ⇒ y = 3 Using this in (ii), we get x = 2y = 2(3) = 6. ∴ The required number = 63. Ans.

Check: The difference of the two digits = 6 - 3 = 3. (verified.) 4 times the number = 4(63) = 252. 7 times the number with the digits reversed = 7(36) = 252 = 4 times the number (verified.)

Exercise on Middle School Math Word Problems

Solve the following Problems from Middle School Math Word Problems.

A two-digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the original number. Find the original number.

The result of dividing a number of two digits by the number with digits reversed is 5⁄6. If the difference of digits is 1, find the number.

If the numerator of a fraction is increased by 2 and its denominator decreased by 1, then it becomes 2⁄3. If the numerator is increased by 1 and denominator increased by 2, then it becomes 1⁄3. Find the fraction.

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